ACIT4330-Page/content/Lectures/Lecture 16.md

lecture, date

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16	2025-03-10

Recap

Recap from the Lecture 15. L_{1}(\mu) = \left\{ \text{measure}\ f : \underbrace{X}_{\mu} \to \mathbb{C} \, | \, \underbrace{\int |f| \, d\mu}_{= \, \sup_{0 \leq s \leq |f|} \int s \, d\mu} \lt \infty \right\} s = \sum_{i = 1}^{n} a_{i} \times X_{A_{i}} \int s \, d\mu = \sum_{i=1}^{n} a_{i} \times \mu(A_{i})

f = \mathrm{Re} f + i \mathrm{Im} f = (\mathrm{Re} f)^{+} - (\mathrm{Re} f)^{-} + i((\mathrm{Im} f)^{+} - (\mathrm{Im} f)^{-}) 0 \leq \int (\mathrm{Re} f)^{\pm} \, d\mu \leq \int |f| \, d\mu

Proposition

L^{1} (\mu) is a Vector Space under Pointwise operations, and the \int : L^{1}(\mu) \to \mathbb{C} \, d\mu is linear, and | \int f \, d\mu | \leq \int |f| \, d\mu.

Proof

Let f, g \in L^{1}(\mu), a \in \mathbb{C}.

Assume first that f, g \geq 0 Simple Function with distinct values a_{i} and b_{i}, let A_{i} = \{ x \in X \, | \, f(x) = a_{i} \} and B_{j} = \{ x \in X \, | \, g(x) = b_{j} \}

Then \int_{A_{i} \cap B_{j}} (f + g) \, d\mu = (a_{i} + b_{j}) \times \mu (A_{i} \cap B_{j}) = a_{i} \times \mu (A_{i} \cap B_{j}) + b_{j} \times \mu(A_{i} + B_{j}) = \int_{A_{i} + B_{j}} f \, d\mu + \int_{A_{i} + B_{j}} g \, d\mu

Note

X \underbrace{=}_{\text{disjoint union}} \cup(A_{i} \cap B_{j}) A \mapsto \int_{A} s \, d\mu Measure L1.

\implies \int_{X} (f + g) \, d\mu = \int f \, d\mu + \int g \, d\mu LHS = \sum_{i j} \int (f + g) \, d\mu = \sum_{i j} \left( \int_{A_{i} \cap B_{j}} f \, d\mu + \int_{A_{i} \cap B_{j}} g \, d\mu\right) = \sum_{i j} \int_{A_{i} \cap B_{j}} f \, d\mu + \sum_{i j} \int_{A_{i} \cap B_{j}} g \, d\mu =^{L_{1}} \int_{X} f \, d\mu + \int_{X} g \, d\mu

L2: 0 \leq s_{1} \leq s_{2} \leq \dots \leq f f=\lim_{ n \to \infty } s_{n}

By L_{2} and Lebesgue's Monotone Convergence Theorem (LMCT) \implies \int is additive for all non-negative Measurable functions f and g. \int (f + g) \, d\mu =^{L_{2}} \int \lim_{ n \to \infty } (s_{n} + s_{n}') \, d\mu =^{\text{LMCT}} \lim_{ n \to \infty } \int (s_{n} + s_{n}') \, d\mu (from above) = \lim_{ n \to \infty } (\int s_{n} \, d\mu + \int s_{n}' \, d\mu (\lim splits) \lim_{ n \to \infty } \int s_{n} \, d\mu + \lim_{ n \to \infty } \int s_{n}' \, d\mu =^{\text{LMCT}} \int f \, d\mu + \int g \, d\mu

For the complex case we have \int \underbrace{| af + g |}_{\leq |a| \times |f| + |g|} \, d\mu \leq \int (| a | \times |f| + |g|) \, d\mu = \int |a| \times |f| \, d\mu + \int |g| \, d\mu = |a| \times \underbrace{\int |f|}_{\lt \infty} \, d\mu + \underbrace{\int |g| \, d\mu}_{\lt \infty} < \infty, so af + g \in L^{1}(\mu) when f, g \in L^{1}(\mu). So L^{1}(\mu) is a Vector Space.

[!info]- Why it's af + g and not af + bg Usually you would use af + bg and have to prove for the bg part as well, but af + g is enough for the proof as you can do:

af + bg = af + (bg + 0)

as the parts in the bracket would belong in L1 as well.

Then use the definition of the integral for complex functions to see that it is additive. Clearly \int a f \, d\mu = a \int f \, d\mu when a \geq 0. For a \lt 0 use (-g)^{\pm} = g^{\mp} etc. Check for a = i, but this is okay as \int i f \, d\mu = \int i (\mathrm{Re} f + i \mathrm{Im} f) \, d\mu = \int (-\mathrm{Im} f + i \mathrm{Re} f) \, d\mu = - \int \mathrm{Im}f \, d\mu + i \int \mathrm{Re} f \, d\mu = i \times \int f \, d\mu. So \int a f \, d\mu = a \int f \, d\mu, \; \forall a \in \mathbb{C}, and the integral is a linear function.

Finally, pick b \in \mathbb{C} such that | \int f \, d\mu | = b \times \underbrace{\int f \, d\mu}_{z}

[!note]- What is happening at z Remember the rules of complex numbers:

|z| = b \times z b = \frac{|z|}{z}

(linear) = \int b f \, d\mu = \int \underbrace{\mathrm{Re}(b f)}_{\leq |b f|} \, d\mu \leq \int |b f| \, d\mu = \int \underbrace{|b|}_{=1} \times |f| \, d\mu = \int |f| \, d\mu QED.

Lebesgue's Dominated Convergence Theorem

It is a theorem for complex valued functions. Also see the Lebesgue's Dominated Convergence Theorem.

Let \{ f_{n} \} be Measurable functions f_{n} : X \to \mathbb{C}, and \mu Measure on X. Assume \exists g \in L^{1} such that all f_{n} \leq g. Then \lim_{ n \to \infty } \int f_{n} \, d\mu = \int \underbrace{\lim_{ n \to \infty } f_{n}}_{f} \, d\mu. If you have a measure so that the whole space is finite, \mu(x) < \infty: \implies \int 1 \, d\mu = \mu(x) < \infty (can use g instead of the 1)

Proof

By Fatou's Lemma \int 2g \, d\mu \underbrace{=}_{f = \lim f_{n}} \int \lim_{ n \to \infty } \inf(\underbrace{2g - |f_{n} - f|}_{\geq 0}) \, d\mu (by Fatou's Lemma) \leq \lim_{ n \to \infty } \inf \int (2g - |f_{n} - f|) \, d\mu (by #Proposition) \int 2g \, d\mu + \lim_{ n \to \infty }\inf\left( -\int |f_{n} - f| \, d\mu \right) = \int 2g \, d\mu - \lim_{ n \to \infty }\sup \int |f_{n} - f | \, d\mu so \lim_{ n \to \infty } \sup \int |f_{n} - f \, d\mu \leq 0, and \lim_{ n \to \infty } \underbrace{\int |f_{n} - f| \, d\mu}_{\geq 0} \leq 0, so \lim_{ n \to \infty } \int |f_{n} - f| \, d\mu = 0.

Then \lim_{ n \to \infty } | \int f_{n} \, d\mu - \int f \, d\mu | (by #Proposition) = \lim_{ n \to \infty } | \int (f_{n} - f) \, d\mu | (again, by #Proposition) \leq \lim_{ n \to \infty } \int |f_{n} - f \, d\mu = 0.

QED.

[!example] Another example

\lim_{ n \to \infty } \lim_{ m \to \infty } \frac{n}{m} = \lim_{ n \to \infty } 0 = 0

Then swap the limits

\lim_{ m \to \infty } \lim_{ n \to \infty } \frac{n}{m} = \lim_{ m \to \infty } \infty = \infty

Which are two different numbers as 0 \neq \infty.